(1) The perimeter of a trapezoid of 10 cm height is 35 cm. If the sum of non-parallel sides is 25 cm,

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1 Grade 8 Mensuration For more such worksheets visit ID : ww-8-mensuration [1] Answer t he quest ions (1) The perimeter of a trapezoid of 10 cm height is 35 cm. If the sum of non-parallel sides is 5 cm, () f ind the area of trapezoid. If the side of the small squares that make up the shape below is 5 cm, then what is the total area of the shape below? (3) Find the surf ace area of a closed cylinder of radius of 7 cm and height 9 cm (Assume π = /7) (4) A wire f rame is bent into a circle of diameter 4 is reshaped as a rhombus. What is the length of the side of the resulting rhombus? Assume π = /7 Choose correct answer(s) f rom given choice (5) If each side of a cube is tripled, f ind the f actor by which surf ace area of the cube will change. a. 9 times b. 7 times c. 3 times d. 1 times (6) A square and an equilateral triangle have the same perimeter. The diagonal of the square is 15 cm. What is the area of the triangle? a. 150 b c. p - 3 d (7) If diagonal of a rhombus are in ratio 5:3, and its area is 70 cm. Find the larger diagonal of the rhombus. a. 8 cm b. 34 cm c. 31 cm d. 30 cm (C) 016 Edugain (

2 (8) T he diagonal of f ollowing quadrilateral ABCD is 40 cm long, and length of perpendiculars dropped on diagonal are 16 cm and 0 cm. Find the area of quadrilateral. ID : ww-8-mensuration [] a. 680 cm b. 70 cm c. 760 cm d. 30 cm (9) One of the diagonal of f ollowing polygon is 60 cm long, and length of perpendiculars dropped on diagonal are 30, 4 cm and 30 cm. Find the area of polygon. a. 981 cm b. 184 cm c. 196 cm d. 0 cm (10) Find area of f ollowing parallelogram (All measurements are in cm). a. 106 cm b. 98 cm c. 100 cm d. 104 cm (11) If diagonals of a rhombus shaped park are 3 and 60 meters, f ind the cost of f encing the park boundary at the rate of $4 per meter. a. 16 b. $544 c. -4 d. $9 (1) How many diagonals will be there in a 11 sides regular polygon? a. 50 b. 44 c. 38 d. 41 (C) 016 Edugain (

3 ID : ww-8-mensuration [3] Fill in the blanks (13) If the diagonal of a square is decreased by 10 %, then the area of the square decreases by %. (14) Barbara f olds a rectangular paper of size 88 cm 5 cm in cylindrical shape such that height of the cylinder is 5 cm. The volume of cylinder = cm 3. (Assume π = /7) (15) If area of f ollowing parallelogram is 31 cm, the height of parallelogram = cm. (All measurements are in cm). 016 Edugain ( All Rights Reserved Many more such worksheets can be generated at (C) 016 Edugain (

4 Answers ID : ww-8-mensuration [4] (1) 50 cm The f ollowing f igure shows the trapezoid ABCD. It is given that the height of the trapezoid ABCD = 10 cm, The perimeter of the trapezoid ABCD = 35 cm. The sum of the non-parallel sides of the trapezoid ABCD = BC + DA = 5 cm. Step The perimeter of the trapezoid ABCD = AB + BC + CD + DA 35 = AB + CD = AB + CD 10 = AB + CD AB + CD = 10 cm The area of the trapezoid ABCD = = = 50 cm AB + CD h Thus, the area of the trapezoid is 50 cm. (C) 016 Edugain (

5 () 800 sq. cm. ID : ww-8-mensuration [5] Since the side of the small squares that make up the given shape is 5 cm. Theref ore the area of the each small square = 5 5 = 5 cm Step If you count the number of small squares that make up the given shape caref ully, you will notice that the number of small squares are 3. Now the area of the shape = 3 5 = 800 sq. cm. (3) 704 cm It is given that the radius (r) and height (h) of cylinder is 7 cm and 9 cm respectively. Step The surf ace area of the closed cylinder = πr(r + h) = 7(7 + 9) 7 = 44(16) = 704 cm Thus, the surf ace area of the closed cylinder is 704 cm. (C) 016 Edugain (

6 (4) 33 ID : ww-8-mensuration [6] A wire f rame of some length was f irst bent into a circle and then reshaped as a rhombus: Wire Circle Rhombus Step Let us f irst f ind the length of the wire f rame. We know that the total length of the boundary of a circle is called its circumf erence and is given by: Circumf erence = πr, where r is the radius of the circle. Since the circle is f ormed by the wire f rame, the length of the wire f rame = πr = 1 [It is given that the radius of the circle is 4/ = 1 and π = /7] 7 = 13 Now, we know that the same wire f rame with length 13 is reshaped as a rhombus. A rhombus has 4 sides and all sides are equal. This means, the length of a side of the rhombus will be 13 divided by 4. That is: 13/4 = 33 Thus the length of the side of the resulting rhombus is 33. (5) a. 9 times Lets assume a is the side of the cube. Surf ace area of the cube = 6a Step According to question each side of a cube is tripled. Now Surf ace area of the cube = (3a) = 9a Now you can say that the f actor by which surf ace area of the cube will change be 9 times. (C) 016 Edugain (

7 (6) b ID : ww-8-mensuration [7] Lets assume s and t are the sides of square and triangle respectively as shown in the f ollowing f igures. Square Equilateral triangle According to question the diagonal AC of the square is 15 cm. Now in right angle triangle ABC AB + BC = AC s + s = (15) (s) = (15) s = (15) s = 11.5 cm Step Perimeter of the given square = 4s Perimeter of the given equilateral triangle = 3t According to question the perimeter of a square is equal to the perimeter of an equilateral triangle. Theref ore 3t = 4s Squaring both sides (3t) = (4s) 9t = 16s t = [Since s = 11.5] 9 t = 00 cm Now the area of an equilateral traingle = t 3 4 = 00 3 [Since t = 00 ] 4 = 50 3 cm Theref ore the area of the triangle is 50 3 cm. (C) 016 Edugain (

8 (7) d. 30 cm ID : ww-8-mensuration [8] It is given that diagonal of a rhombus are in ratio 5:3. Theref ore, let us assume that the diagonals of rhombus are 5x cm and 3x cm. Step Area of the rhombus = (Multiplication of diagonals) / 70 = (5x)(3y) 70 = 15x x = 70 / 15 x = 36 x = 6... (Since x cannot be negative) T heref ore, two diagonals are, 5 x = 5 6 = 30 cm 3 x = 3 6 = 18 cm Hence larger diagonal of the rhombus is 30 cm. (C) 016 Edugain (

9 (8) b. 70 cm ID : ww-8-mensuration [9] Given, Length of the diagonal, AC = 40 cm Length of the perpendiculars, BE and DF are 16 cm and 0 cm Step We know that the area of a triangle = Base Height Area of the triangle ABC = AC BE = = 30 cm Area of the triangle ACD = AC DF = 40 0 = 400 cm Thus, the area of the quadrilateral ABCD = Area of the triangle ABC + Area of the triangle ACD = = 70 cm (C) 016 Edugain (

10 (9) c. 196 cm ID : ww-8-mensuration [10] We know that, the area of a triangle = 'The base of the triangle' 'The height of the triangle' The sum of the lengths of the parallel sides The area of a trapezium = the trapezium' Step The f ollowing f igure shows the required polygon, 'The height of The area of the polygon = The area of the ΔABC + The area of the ΔAGD + The area of the ΔEFC + The area of the trapezium DEFG = = = 196 cm Thus, the area of the polygon is 196 cm. + (30+4) 18 (C) 016 Edugain (

11 (10) d. 104 cm ID : ww-8-mensuration [11] If we look at the f igure, we notice that, the base of the parallelogram = 13 cm, and the height of the parallelogram = 8 cm. Step The area of the parallelogram = Base Height = 13 8 = 104 cm (C) 016 Edugain (

12 (11) b. $544 ID : ww-8-mensuration [1] The f ollowing f igure shows the rhombus shaped park, Step It is given that, the diagonal of the rhombus shaped park are 3 and 60 meters. Theref ore, AC = 3 meters, BD = 60 meters We know that the diagonals of a rhombus bisect each other at right angles. Theref ore, the OC = AC/ = 3/ = 16 meters, OD = BD/ = 60/ = 30 meters, and COD = 90 Now, in right angled triangle COD, CD = (OC + OD ) = ( ) = = 1156 = 34 meters We know that the all sides of a rhombus are congruent. The perimeter of the rhombus ABCD = 4(CD) = 4 34 = 136 meters Step 5 The cost of f encing the one meter park boundary = $4 The cost of f encing the 136 meters park boundary = = $544 Step 6 Thus, the cost of f encing the park boundary is $544. (C) 016 Edugain (

13 (1) b. 44 ID : ww-8-mensuration [13] A Polygon's diagonals are line segments f rom one corner to another, but not the sides. The number of diagonals of an n-sided polygon is given by n(n-3). Step Number of sides of the regular polygon in question, n = 11 Number of diagonals = n(n-3) = 11(11-3) = 88 = 44 Theref ore, the number of diagonals in a 11 sides regular polygon will be 44. (C) 016 Edugain (

14 (13) 19 ID : ww-8-mensuration [14] Let the length of the diagonal of the square be d. The length of the side of the square will then be d /, and the area of the square will be (d / ) (d / ) = 0.5d Step Af ter reducing the length of the diagonal by 10%, the new length of the diagonal will be: 10 = d - d 100 = 0.9d Hence the new area will be 0.5(0.9d) = d. The decrease in area = Old Area - New Area = 0.5 d d = 0.5 (1-0.81) d = d Step 5 Percentage decrease in area = Decrease in area Old Area 100 % d = 100 % 0.5 d = % = 19% Step 6 Hence, when the diagonal of a square is decreased by 10%, then the area of the square decreases by 19%. (C) 016 Edugain (

15 (14) 3080 ID : ww-8-mensuration [15] The paper has to be f olded along the length in order to get the cylinder of height 5 cm and the perimeter of the base of the cylinder is equal to the length of the paper. Step Let the radius of cylinder be r cm. Since, the base of the cylinder is circular, its perimeter is πr. Theref ore, πr = 88 /7 r = 88 44/7 r = 88 r = 88 7/44 r = 14 cm The radius of the cylinder = 14 cm Now, the volume of the cylinder = πr h = = 3080 cm 3 (15) 1 If we look at the f igure, we notice that, the base the parallelogram is 6 cm. Step It is given that, the area of the parallelogram is 31 cm. We know that, the area of the parallelogram = The base of the parallelogram The height of the parallelogram Theref ore, 31 = 6 The height of the parallelogram 31 = The height of the parallelogram 6 1 = The height of the parallelogram The height of the parallelogram = 1 cm Thus, the height of the parallelogram is 1 cm. (C) 016 Edugain (